Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

PhenotypeGenotype Genes code for proteins (or RNA). These gene products give rise to traits… It is rarely this simple.

Epigenetics: Lamarck was right? Sort of… Image from:

Genomic Imprinting Genomic imprinting is a phenomenon in which expression of a gene depends on whether it is inherited from the male or the female parent Imprinted genes follow a non-Mendelian pattern of inheritance –Depending on how the genes are “marked”, the offspring expresses either the maternally- inherited or the paternally-inherited allele ** Not both

A hypothetical example of imprinting A=curly hair a=straight hair B=beady eyes b=normal *=methylation A* in males B* in females A*a bB* A*a bB* A*a bB Aa bB* A*b, A*B, ab, aB Ab, AB*, ab, aB* a B* a B* A* b A* b similar to Fig 7.10

Genomic imprinting must involve a marking process At the molecular level, the imprinting is known to involve differentially methylated regions –They are methylated either in the oocyte or sperm Not both Imprinting and DNA Methylation

For most genes, methylation results in inhibition of gene expression –However, this is not always the case

Haploid female gametes transmit an unmethylated gene Haploid male gametes transmit a methylated gene Fig 7.11 Changes in methylation during gamete development alter the imprint

Thus genomic imprinting is permanent in the somatic cells of an animal – However, the marking of alleles can be altered from generation to generation

To date, imprinting has been identified in dozens of mammalian genes Tbl 7.2

Imprinting plays a role in the inheritance of some human diseases: Prader-Willi syndrome (PWS) and Angelman syndrome (AS) –PWS is characterized by: reduced motor function, obesity, mental deficiencies –AS is characterized by: hyperactivity, unusual seizures, repetitive muscle movements, mental deficiencies Usually, PWS and AS involve a small deletion in chromosome 15 –If it is inherited from the mother, it leads to AS –If it is inherited from the father, it leads to PWS

AS results from the lack of expression of UBE3A (encodes a protein called EA-6P that transfers small ubiquitin molecules to certain proteins to target their degradation) –The gene is paternally imprinted (silenced) PWS results (most likely) from the lack of expression of SNRNP (encodes a small nuclear ribonucleoprotein that controls gene splicing necessary for the synthesis of critical proteins in the brain) –The gene is maternally imprinted (silenced)

Fig 7.12 The deletion is the same in males and females, but the expression is different depending on who you received the normal version from.

The relationship between genes and traits is often complex Complexities include: Complex relationships between alleles

The relationship between genes and traits is often complex Complexities include: Multiple genes controlling one trait

Two genes control coat color in mice Fig 4.21

Variation in Peas Fig 3.2

Fig 2.8 Inheritance of 2 independent genes

Y y rR Gene for seed color Gene for seed shape Approximate position of seed color and shape genes in peas Chrom. 1/7Chrom. 7/7

There must be a better way… Fig 2.9

Inheritance can be predicted by probability Section 2.2, pg 30-32

Sum rule The probability that one of two or more mutually exclusive events will occur is the sum of their respective probabilities Consider the following example in mice Gene affecting the ears –De = Normal allele –de = Droopy ears Gene affecting the tail –Ct = Normal allele –ct = Crinkly tail

If two heterozygous (Dede Ctct) mice are crossed Then the predicted ratio of offspring is –9 with normal ears and normal tails –3 with normal ears and crinkly tails –3 with droopy ears and normal tails –1 with droopy ears and crinkly tail These four phenotypes are mutually exclusive –A mouse with droopy ears and a normal tail cannot have normal ears and a crinkly tail Question –What is the probability that an offspring of the above cross will have normal ears and a normal tail or have droopy ears and a crinkly tail?

Applying the sum rule –Step 1: Calculate the individual probabilities 9( )= 9/16P (normal ears and a normal tail) = 1( )= 1/16P (droopy ears and crinkly tail) = –Step 2: Add the individual probabilities 9/16 + 1/16 = 10/16 10/16 can be converted to –Therefore 62.5% of the offspring are predicted to have normal ears and a normal tail or droopy ears and a crinkly tail

Product rule The probability that two or more independent events will occur is equal to the product of their respective probabilities Note –Independent events are those in which the occurrence of one does not affect the probability of another

Consider the disease congenital analgesia –Recessive trait in humans –Affected individuals can distinguish between sensations However, extreme sensations are not perceived as painful –Two alleles P = Normal allele p = Congenital analgesia Question –Two heterozygous individuals plan to start a family –What is the probability that the couple’s first three children will all have congenital analgesia?

Applying the product rule –Step 1: Calculate the individual probabilities This can be obtained via a Punnett square 1/4P (congenital analgesia) = –Step 2: Multiply the individual probabilities 1/4 X 1/4 X 1/4 = 1/64 1/64 can be converted to –Therefore 1.6% of the time, the first three offspring of a heterozygous couple, will all have congenital analgesia

Crossing- over Meiosis I Meiosis II 4 Haploid cells, each unique (Ind. Assort.) Different genes are not always independent

The haploid cells contain the same combination of alleles as the original chromosomes The arrangement of linked alleles has not been altered Fig 5.1

These haploid cells contain a combination of alleles NOT found in the original chromosomes These are termed parental or non- recombinant cells This new combination of alleles is a result of genetic recombination These are termed recombinant cells Fig 5.1

Linked alleles tend to be inherited together

Recombinants are produced by crossing over

Crossing over produces new allelic combinations

Only 2 chromosomes cross-over, and so the maximum number of recombinants that can be produced is 50%.

For linked genes, recombinant frequencies are less than 50 percent

Homologous pair of chromosomes

Does this show recombination? D/d M1/M2 d/d M1/M2 D/d M1/M2 d/d M2/M2 D/d M2/M2 d/d M2/M2

Does this show recombination? D/d M1/M2 d/d M1/M2 D/d M1/M2 d/d M2/M2 D/d M2/M2 d/d M2/M2

Longer regions have more crossovers and thus higher recombinant frequencies

Some crosses do not give the expected results

=25% 8%9%41%42%

These two genes are on the same chromosome

By comparing recombination frequencies, a linkage map can be constructed

= 17 m.u.

Linkage map of Drosophila chromosome 2: This type of map, with mapping units more than 50, can only be put together by making comparisons of linked genes.

The probability of crossing over can be used to determine the spatial relationship of different genes

similar to Fig 5.3, also see Fig 5.9, and pg What is the relationship between these 3 genes? What order and how far apart?